■トリボナッチ数列(その25)
3次方程式の解を-1、ω、ω^-1とする。t0=1,t1=135,t2=11161,t3=926271
ω+ω^-1=83
ω・ω^-1=1
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α=-1
β=ω
γ=ω^-1
α+β+γ=-1+ω+ω^-1=82
αβ+βγ+γα=1-ω+ω^-1=-82
αβγ=-1
Un+1=Tn+1-αTn
U0=T1-αT0=136
U1=T2-αT1=11296
U2=T3-αT2=937432
Vn=Un+1-βUn
V0=U1-βU0=11296-136ω
V1=U2-βU1=937432-ω(11296-136ω)
Vn+1=γVn
Vn=ω^-n・(11296-136ω)
Wn=Un+1-γUn
W0=U1-γU0=11296-136ω^-1
W1=U2-γU1=937432-ω^-1(11296-136ω^-1)
Wn+1=βWn
Wn=ω^n・(11296-136ω^-1)
Un+1-βUn=ω^-n・(11296-136ω)
Un+1-γUn=ω^n・(11296-136ω^-1)
Un=(ω^n・(11296-136ω^-1)-ω^-n・(11296-136ω))/(ω-ω^-1)
Tn+1-αTn=(ω^n・(11296-136ω^-1)-ω^-n・(11296-136ω))/(ω-ω^-1)
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t0=1,t1=135,t2=11161,t3=926271
Un+1=Tn+1-βTn
U0=T1-βT0=135-ω
U1=T2-βT1=11161-135ω
U2=T3-βT2=926271-11161ω
Vn=Un+1-γUn
V0=U1-γU0=11161-135ω-ω^-1・(135-ω)=-43
V1=U2-γU1
Vn+1=αVn
Vn=(-1)^n・(-43)
Wn=Un+1-αUn
W0=U1-αU0=11161-135ω+(135-ω)=11296-136ω
W1=U2-αU1=
Wn+1=γWn
Wn=ω^-n・(11296-136ω)
Un+1-γUn=(-1)^n・(-43)
Un+1-αUn=ω^-n・(11296-136ω)
Un=(ω^-n・(11296-136ω)-(-1)^n(-43))/(ω^-1+1)
Tn+1-βTn=(ω^-n・(11296-136ω)-(-1)^n(-43)))/(ω^-1+1)
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t0=1,t1=135,t2=11161,t3=926271
Un+1=Tn+1-γTn
U0=T1-γT0=135-ω^-1
U1=T2-γT1=11161-135ω^-1
U2=T3-γT2=926271-11161ω^-1
Vn=Un+1-βUn
V0=U1-βU0=11161-135ω^-1-ω・(135-ω^-1)=-43
V1=U2-βγU1
Vn+1=αVn
Vn=(-1)^n・(-43)
Wn=Un+1-αUn
W0=U1-αU0=11161-135ω^-1+(135-ω^-1)=11296-136ω^-1
W1=U2-αU1=
Wn+1=βWn
Wn=ω^n・(11296-136ω^-1)
Un+1-βUn=(-1)^n・(-43)
Un+1-αUn=ω^n・(11296-136ω^-1)
Un=(ω^n・(11296-136ω^-1)-(-1)^n(-43))/(ω+1)
Tn+1-γTn=(ω^n・(11296-136ω^-1)-(-1)^n(-43)))/(ω+1)
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Tn=-{(β-γ)α^n+(γ-α)β^n+(α-β)γ^n}/(α-β)(β-γ)(γ-α)
α^n→(-1)^n・(-43)
β^n→ω^n・(11296-136ω^-1)
γ^n→ω^-n・(11296-136ω)に置換
(α-β)=-1-ω
(β-γ)=(ω-ω^-1)
(γ-α)=(ω^-1+1)
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n=0のとき
T0=-{(β-γ)(-43)+(γ-α)(11296-136ω^-1)+(α-β)(11296-136ω)}/(α-β)(β-γ)(γ-α)=1になるだろうか?
もっと簡単になればよいのであるが・・・x^2-83x+1=0
x=1/2・{83+-{83^2-4}^1/2}
x=1/2・{83+-6885^1/2}
x=1/2・{83+-82.9759・・・}
(ω-ω^-1)=(6885)^1/2
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